Characterizing Implicitly Defined Function (asymptotically)?

Question

I would like to characterize an implicitly defined function $m_n(X)$ that satisfies $$ e^{-b m_n(X)} \sum_{i=0}^{n-1}\frac{(b m_n(X))^i}{i!} = e^{-X}. $$ In particular, I want to know how $m_n(X)$ depends on $n$, which is discrete and ranges from $\{1, 2, ..., N\}$. For example, it would be useful for my study to know if $$ m_n(X) = \frac{X}{b} + (n - 1) \frac{\ln X}{b} + o\left( n \frac{\ln X}{b} \right), $$ though I am not sure whether this is true; we can note that $$ e^{-b m_n(X)} \frac{(b m_n(X))^{n-1}}{(n-1)!} \sim e^{-X}, $$ but I'm not sure where to go from there to prove the result. Any other useful thoughts on characterizing how $m_n(X)$ varies with $n$ would be greatly appreciated -- especially if there is a useful way to proceed without resorting to asymptotics.

Answer 1

I will assume that $X$ and $bm_n(X)$ are large and $n$ is bounded (or at least $n \ll bm_n(X)$). In terms of the normalised incomplete gamma function $Q$, your equation is $$ Q(n,bm_n (X)) = \mathrm{e}^{ - X} . $$ It is well known that $$ Q(n,z) \sim \frac{{\mathrm{e}^{ - z} z^{n - 1} }}{{(n - 1)!}}\left( {1 + \frac{{n - 1}}{z} + \cdots } \right) $$ as $z\to +\infty$. Consequently, $$ \frac{1}{{Q(n,z)}} \sim \frac{{\mathrm{e}^z }}{{z^{n - 1} }}\left( {(n - 1)! - \frac{{(n - 1)(n - 1)!}}{z} + \ldots } \right) $$ as $z\to +\infty$. Take $\mathrm{e}^{ - X} = Q(n,z)$. By a result of G. Robin (see, e.g., here), we can invert the above asymptotics to deduce $$ z = X + (n - 1)\log X - \log ((n - 1)!) + \mathcal{O}\!\left( {n^2 \frac{{\log X}}{X}} \right). $$ Accordingly, $$ m_n (X) = \frac{X}{b} + (n - 1)\frac{{\log X}}{b} - \frac{{\log ((n - 1)!)}}{b} + \mathcal{O}\!\left( {n^2 \frac{{\log X}}{{bX}}} \right). $$ For asymptotic inversions of $Q(n,z)$ that are more uniform in $n$ and $z$, see this and this paper.